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Furnishing a contradiction in my proof involving Lagrange's Theorem

  • Thread starter jdinatale
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  • #1
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Homework Statement


furnish.jpg



The Attempt at a Solution


I'm trying to show that Case 1 implies that [itex]\tau \in H[/itex], and since [itex]\tau[/itex] was an arbitrarily chosen 3-cycle, then H must contain all 3-cycles, thus contradicting that H has 6 elements.
 

Answers and Replies

  • #2
Deveno
Science Advisor
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first of all, what is σ? it appears to be an arbitrary 3-cycle, but that isn't clear.

but going on that assumption, note that one of the elements of H is e.

therefore σ is in Hσ. similarly σσ is in Hσσ.

but since Hσ = Hσσ, σσ = τkσ, so σ = σσσσ = τkσσσ = τk.

thus σ is in H, which means <σ> = {σ,σσ,e} is in H.

but σ is an arbitrary 3-cycle, so....
 

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